🗺️ IQ5 · Lesson 19 of 23 · 50 min

Organic Reaction Pathways — Synthesis & Multi-Step Problems

Reaction pathways are the HSC exam skill that separates Band 4 from Band 6 — not because the individual steps are hard, but because you need every reaction from L06 to L18 working together simultaneously, and one missing condition or wrong arrow invalidates the whole chain.

Printable worksheet

Download this lesson's worksheet

Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.

Download PDF Open printable version

🗺️ Think First — Before You Read

A food scientist needs to make banana flavouring (ethyl ethanoate) in three steps from ethene and ethanol. A pharmaceutical chemist needs to synthesise an ester from a haloalkane starting material in four steps.

Before reading: draw the functional group changes needed to get from ethene (CH₂=CH₂) to ethyl ethanoate (CH₃COOC₂H₅). How many steps? What changes at each step?

Complete Reaction Conditions Reference

ALKENE → ALCOHOL: + H₂O (steam), H₃PO₄ cat., ~300°C, high pressure

ALKENE → HALOALKANE: + HX, no cat., r.t. (Markovnikov — X to more substituted C)

ALKANE → HALOALKANE: + X₂, UV light, r.t.

HALOALKANE → ALCOHOL: + NaOH(aq), reflux

PRIMARY ALCOHOL → ALDEHYDE: + K₂Cr₂O₇/H₂SO₄, distillation (remove aldehyde before over-oxidation)

PRIMARY ALCOHOL → CARBOXYLIC ACID: + K₂Cr₂O₇/H₂SO₄ (excess), reflux

SECONDARY ALCOHOL → KETONE: + K₂Cr₂O₇/H₂SO₄, reflux (DEAD END — ketone cannot be further oxidised)

CARBOXYLIC ACID + ALCOHOL ⇌ ESTER + H₂O: conc. H₂SO₄ cat., reflux

ESTER + NaOH → CARBOXYLATE + ALCOHOL: conc. NaOH, reflux, → (saponification, irreversible)

KEY: Distillation = stop at aldehyde | Reflux + excess oxidant = go to carboxylic acid

Know

All 15+ reaction types and their conditions from L06–L18
Which functional group conversions are possible vs impossible in Module 7
The Markovnikov limitation in hydration of asymmetrical alkenes

Understand

Why distillation vs reflux determines aldehyde vs carboxylic acid
Why secondary alcohol oxidation is a dead end (ketone)
How to choose the shortest valid path between two functional groups

Can Do

Apply the four-step algorithm to any pathway problem
Write balanced equations with full conditions for each step
Construct a four-step synthesis from haloalkane to ester

Key Terms — scan these before reading

HydrocarbonAn organic compound containing only carbon and hydrogen atoms.

Functional groupA specific atom arrangement determining characteristic chemical reactions.

Homologous seriesA family of compounds with the same functional group, differing by CH₂.

Addition polymerA polymer formed by monomers adding together without loss of atoms.

Condensation polymerA polymer formed with elimination of a small molecule such as water.

EsterificationA condensation reaction between a carboxylic acid and an alcohol forming an ester.

Core Concepts

1 — The Complete Reaction Map: Every Arrow You Have Built

By L19 you have learned 15+ individual reaction types. This card assembles them into a single connected map so you can see which functional groups connect to which and what the shortest path between any two points is.

HSC Must-Do: Print or draw this reaction map and keep it visible during all Module 7 revision. When you encounter a pathway question, trace the route on the map FIRST before writing any equations. The map shows all possible routes and lets you choose the shortest one. Attempting pathway questions without a visual map is the most common cause of incomplete answers.

Common Error: Students draw arrows that do not exist — for example, alkane → alcohol directly (no such route in Module 7 — must go alkane → haloalkane → alcohol). Or ketone → carboxylic acid (impossible — ketones cannot be oxidised further). Every arrow must have known conditions behind it.

Exam TipFor organic chemistry questions, draw full structural formulas showing all atoms and bonds — condensed or skeletal formulas alone may lose marks in HSC extended-response questions.

2 — The Systematic Approach: Four-Step Algorithm for Pathway Problems

A multi-step pathway problem is not solved by intuition — it is solved by a systematic algorithm: identify functional groups of start and end, find the shortest connected path on the map, then write each step with full conditions.

Identify functional groups. Look at the starting material and target product. Write down the functional group class of each, and note if the carbon skeleton changes (if it does — the pathway is invalid for Module 7 reactions that don't alter chain length).

Locate on the reaction map. Find the start and target functional groups on the map. Count the minimum number of arrows between them. Each arrow = one step = one set of conditions.

Check each step. Can this transformation be done in one reaction? Are there Markovnikov/anti-Markovnikov issues? Does distillation vs reflux matter (aldehyde vs carboxylic acid)? Is there a dead end (secondary alcohol → ketone)?

Write equations with full conditions. For each step: (1) balanced structural equation; (2) conditions box — REAGENT / CATALYST / EQUIPMENT; (3) name of the intermediate compound produced.

The Critical Distillation vs Reflux Decision (primary alcohol oxidation):

Primary Alcohol
R–CH₂OH

→
K₂Cr₂O₇
distillation

Aldehyde
R–CHO

→
K₂Cr₂O₇ excess
reflux

Carboxylic Acid
R–COOH

Secondary Alcohol
R–CHOH–R'

→
K₂Cr₂O₇
reflux

Ketone ⚠
R–CO–R'

Cannot oxidise further — dead end in Module 7

Trap 1 — Oxidation levelAldehyde and carboxylic acid both come from primary alcohol — distillation gives aldehyde; reflux + excess gives acid. Specify which every time.

Trap 2 — Alkane to alcoholAlkane → alcohol is NOT a direct Module 7 reaction. Must go: alkane → haloalkane → alcohol (two steps).

Trap 3 — MarkovnikovHydration or hydrohalogenation of asymmetrical alkene puts functional group at more substituted carbon (C2, C3...). Check if the target has the OH at C1 — if so, Markovnikov is the wrong approach.

Trap 4 — Esterification arrowEsterification is REVERSIBLE (⇌). Always write ⇌, state conc. H₂SO₄ catalyst and reflux, and note yield <100%.

HSC Must-Do: For every pathway step, write the name AND structural formula of each intermediate — not just the final product. The marker needs to see that you know what is produced at each stage. Skipping intermediates earns zero marks for those steps even if the final product is correct.

Common Error: Students write steps in the wrong order — they work backwards in their heads but write forward in the wrong sequence. Always write forward: Step 1 → Intermediate 1; Step 2 → Intermediate 2; ... → Target. Number each step clearly.

3 — Pathway Drills: Building From Simple to Complex

The best way to internalise the algorithm is to apply it to progressively more complex examples — starting with two-step paths and building toward four-step syntheses.

Drill 1 — Ethanol → Ethyl Ethanoate (2 steps, same starting material in both roles):

Step 1

Equation: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O

Conditions: K₂Cr₂O₇/H₂SO₄ (excess), heat, reflux. Orange → green.

Intermediate: Ethanoic acid

Step 2

Equation: CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O

Conditions: Conc. H₂SO₄ (catalyst), heat under reflux. Reversible — yield ~65%.

Intermediate: Ethyl ethanoate

Note: Ethanol is used in two roles — oxidise part to ethanoic acid, keep the rest for esterification. This "split the starting material" strategy is common in HSC pathway questions.

Drill 2 — Propan-1-ol → Propyl Propanoate (3 steps, same starting material in two roles):

Step 1

Equation: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O

Conditions: K₂Cr₂O₇/H₂SO₄, distillation (collect propanal before over-oxidation).

Intermediate: Propanal

Step 2

Equation: CH₃CH₂CHO + [O] → CH₃CH₂COOH

Conditions: K₂Cr₂O₇/H₂SO₄ (excess), reflux.

Intermediate: Propanoic acid

Step 3

Equation: CH₃CH₂COOH + CH₃CH₂CH₂OH ⇌ CH₃CH₂COOC₃H₇ + H₂O

Conditions: Conc. H₂SO₄ catalyst, reflux. Reversible (⇌).

Intermediate: Propyl propanoate

HSC Must-Do: When a pathway says "starting from compound X only," you may need to use it in two different roles — for example, oxidise part to a carboxylic acid, then esterify another portion of the original alcohol with that acid. This is chemically valid and expected.

Common Error: Reversing oxidation steps — writing "aldehyde + K₂Cr₂O₇ → alcohol" (backwards: this is reduction, not in Module 7). Check the oxidation state of each intermediate: alcohol (C-OH) < aldehyde (C=O, 1H) < carboxylic acid (COOH). You can only go uphill (oxidise) in Module 7.

4 — Complex Pathway Problems: Four-Step Syntheses and Unknown Starting Materials

The hardest pathway questions give you an unfamiliar starting material and expect you to use functional group identification + the reaction map to construct a valid route — this card builds that skill explicitly.

Approach for unknown compounds: ALWAYS identify the functional group class of the unknown first. From the functional group, locate it on the map, then trace the path.

Four-Step Worked Pathway — 1-Bromobutane → Butyl Butanoate:

Step	Equation	Conditions	Why these conditions?
Step 1	CH₃CH₂CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaBr	NaOH(aq), reflux	Aqueous NaOH → substitution → alcohol. Alcoholic NaOH → elimination → alkene instead.
Step 2	CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O	K₂Cr₂O₇/H₂SO₄, distillation	Butanal (BP 75°C) is more volatile than butan-1-ol (BP 118°C) — distilling removes aldehyde before excess oxidant converts it to carboxylic acid.
Step 3	CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH	K₂Cr₂O₇/H₂SO₄ (excess), reflux	Reflux keeps everything in the flask — excess oxidant completes the conversion to carboxylic acid. No distillation needed here.
Step 4	CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂O	Conc. H₂SO₄ (catalyst), reflux, reversible (⇌)	H₂SO₄ activates acid and dehydrates water produced (Le Chatelier → shift right → higher yield). Butan-1-ol from Step 1 is used here.

HSC Must-Do: In every multi-step pathway answer, present steps as a numbered sequence with: (1) the reaction equation; (2) conditions box (REAGENT/CATALYST/EQUIPMENT); (3) the name of the intermediate or final product. A paragraph without numbered steps is very difficult for the marker to follow, and risks losing partial-mark credit.

Common Error: Carbon chain length changing unexpectedly. Every Module 7 functional group interconversion preserves the carbon skeleton of the molecule. The only step where two molecules combine is esterification (and amide formation) — and the only step where one molecule splits is saponification. Do not draw pathways where a 4-carbon chain suddenly becomes 3 carbons without explanation.

⚠ Common Misconceptions — Pathway Problems

Misconception 1: "I can go directly from alkane to alcohol." Reality: No Module 7 reaction converts alkane → alcohol directly. Must go alkane → haloalkane (X₂/UV) → alcohol (NaOH(aq)/reflux) — two steps.

Misconception 2: "Reflux always gives an aldehyde." Reality: Reflux with excess oxidant gives a carboxylic acid. Distillation gives an aldehyde. Equipment determines the product.

Misconception 3: "A secondary alcohol can be oxidised to a carboxylic acid." Reality: Secondary alcohol → ketone only. Ketones CANNOT be oxidised further to carboxylic acids under Module 7 conditions — dead end.

Misconception 4: "Esterification is irreversible." Reality: Esterification is REVERSIBLE (⇌). Must write ⇌ in the equation. Saponification with NaOH is irreversible (→).

Worked Examples

Example 1 — Two-Step Pathway Identification (Straightforward)

Problem: Identify the reagents and conditions for each step. Name all intermediates.
Ethanol → [Step 1] → Intermediate A → [Step 2] → Ethyl ethanoate

Given: Starting material: ethanol. Target: ethyl ethanoate (CH₃COOC₂H₅). Two steps.

Find: Identify Intermediate A and conditions for both steps.

Identify Intermediate A: Ethyl ethanoate = ethyl (from ethanol) + ethanoate (from ethanoic acid). Esterification needs a carboxylic acid. Therefore Intermediate A = ethanoic acid (CH₃COOH).

Step 1 — ethanol → ethanoic acid:
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
Conditions: K₂Cr₂O₇/H₂SO₄ (excess), heat under reflux. Colour change orange → green.

Step 2 — esterification:
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O
Conditions: conc. H₂SO₄ (catalyst), heat under reflux. Reversible (⇌), yield ~65%.

Intermediate A = ethanoic acid. Step 1: excess K₂Cr₂O₇/H₂SO₄, reflux. Step 2: conc. H₂SO₄ cat., reflux, ⇌.

Example 2 — Three Parallel Pathways from 1-Bromopropane (Intermediate)

Problem: Starting from 1-bromopropane (CH₃CH₂CH₂Br), outline: (a) One-step synthesis of propan-1-ol. (b) Two-step synthesis of propanal. (c) Two-step synthesis of propanoic acid.

(a) 1-bromopropane → propan-1-ol (1 step):
CH₃CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂OH + NaBr
Conditions: NaOH(aq), reflux. Intermediate: propan-1-ol.

(b) → propanal (2 steps):
Step 1: same as (a) → propan-1-ol.
Step 2: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O
Conditions: K₂Cr₂O₇/H₂SO₄, distillation. Intermediate: propanal (aldehyde).

(c) → propanoic acid (2 steps):
Step 1: same as (a) → propan-1-ol.
Step 2: CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O
Conditions: K₂Cr₂O₇/H₂SO₄ (excess), reflux. Product: propanoic acid.

(a) NaOH(aq), reflux → propan-1-ol. (b) then K₂Cr₂O₇/distillation → propanal. (c) then K₂Cr₂O₇ excess/reflux → propanoic acid. Key distinction: distillation stops at aldehyde; reflux + excess goes to acid.

Example 3 — Extended Four-Step Synthesis (Hard, 8 marks)

Problem: Starting from 1-bromobutane (CH₃CH₂CH₂CH₂Br), describe a four-step synthesis of butyl butanoate (CH₃CH₂CH₂COOC₄H₉). For each step write the balanced equation, all conditions, name the intermediate, and explain why those conditions give the desired intermediate rather than an alternative product.

Step 1 — 1-bromobutane → butan-1-ol:
CH₃CH₂CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaBr
Conditions: NaOH(aq), reflux. Intermediate: butan-1-ol.
Why aqueous NaOH: Aqueous NaOH provides OH⁻ for nucleophilic substitution at the C-Br bond → alcohol. Alcoholic (ethanolic) NaOH would instead promote elimination (E2) → alkene — wrong product.

Step 2 — butan-1-ol → butanal:
CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O
Conditions: K₂Cr₂O₇/H₂SO₄, distillation. Intermediate: butanal.
Why distillation: Butanal (BP 75°C) has a lower boiling point than butan-1-ol (BP 118°C). Distilling removes butanal as it forms, preventing further oxidation to butanoic acid by the excess dichromate. Without distillation (reflux), butanoic acid would be the product.

Step 3 — butanal → butanoic acid:
CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH
Conditions: K₂Cr₂O₇/H₂SO₄ (excess), reflux. Intermediate: butanoic acid.
Why reflux: Reflux keeps the aldehyde in contact with excess oxidant until complete oxidation to the carboxylic acid occurs. Distillation here would remove the aldehyde prematurely — we want full oxidation, so the reflux condenser returns vapour to the flask.

Step 4 — butanoic acid + butan-1-ol → butyl butanoate:
CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂O
Conditions: conc. H₂SO₄ (catalyst), heat under reflux. Reversible (⇌).
Why conc. H₂SO₄: H₂SO₄ provides H⁺ to activate the carboxylic acid for nucleophilic attack by the alcohol. It is regenerated (true catalyst). It also acts as a dehydrating agent — absorbing water produced, shifting equilibrium right by Le Chatelier's principle → higher ester yield.
Note: Butan-1-ol from Step 1 serves both as the starting material for Steps 2-3 AND as the alcohol for Step 4 — produce sufficient butan-1-ol, split the batch.

Transformation	Reagent/Catalyst	Equipment	Key point
Alkene → Alcohol	H₂O, H₃PO₄ cat.	High T (300°C), high P	Markovnikov — OH to more substituted C
Alcohol → Alkene	Conc. H₂SO₄ or H₃PO₄	Heat (~170°C)	Reverse of hydration
Alkane → Haloalkane	X₂, UV light	Room temp.	Radical substitution; mixture of products
Haloalkane → Alcohol	NaOH(aq)	Reflux	Aqueous = substitution (not elimination)
Primary alcohol → Aldehyde	K₂Cr₂O₇/H₂SO₄	Distillation	Remove product to prevent over-oxidation
Primary alcohol → Carboxylic acid	K₂Cr₂O₇/H₂SO₄ (excess)	Reflux	Keep in flask for full oxidation
Secondary alcohol → Ketone	K₂Cr₂O₇/H₂SO₄	Reflux	DEAD END — cannot oxidise further
Acid + Alcohol ⇌ Ester + H₂O	Conc. H₂SO₄ (cat.)	Reflux	Reversible; H₂SO₄ also dehydrates
Ester + NaOH → Soap + Alcohol	NaOH(aq)	Reflux	Saponification — IRREVERSIBLE (→)
RCOOH + RNH₂ → Amide + H₂O	Heat	Condensation	Amide bond = peptide bond

Activities

Activity 1 — Pathway Map Practice

Using only Module 7 reactions, plan the shortest valid pathway for each of the following. State the number of steps and the conditions for each step:

Ethene → ethanoic acid
1-bromopropane → propyl propanoate
Propan-2-ol → propanone (1 step)
But-1-ene → butanal (is this achievable? — explain the challenge with Markovnikov)

Activity 2 — Spot the Error

Each of the following pathway steps contains one error. Identify and correct each:

"Butan-2-ol + K₂Cr₂O₇/H₂SO₄, excess, reflux → butanoic acid"
"Ethene + NaOH(aq) → ethanol"
"Propan-1-ol + K₂Cr₂O₇/H₂SO₄, reflux → propanal"
"Ethanoic acid + ethanol ⇌ ethyl ethanoate + H₂O (no catalyst needed)"

Interactive

Check Your Understanding — Multiple Choice

Q1. A student wants to synthesise ethyl propanoate. Which combination of starting materials and pathway is correct?

Q2. In a multi-step synthesis, a student needs to convert a primary alcohol to an aldehyde (not a carboxylic acid). Which combination of oxidising agent and equipment is correct?

Q3. Which of the following is NOT a valid two-step pathway within Module 7 scope?

Q4. A student wants to convert 1-chloropropane to propanoic acid in three steps. What is the correct sequence?

Q5. Which statement best explains why conc. H₂SO₄ is used as a catalyst in esterification rather than being a reactant?

Short Answer Questions

Q6. (4 marks) Outline a two-step synthesis of ethyl ethanoate starting from ethene only. For each step, write the balanced equation, state all conditions, and name the compound produced.

Q7. (5 marks) Starting from 1-bromobutane, outline a four-step synthesis of butyl butanoate. For each step, write the balanced equation, conditions, and name the intermediate. Explain why distillation is used in one step and reflux in another during the oxidation sequence.

Q8. (6 marks) A student proposes the following pathway: "butan-2-ol → butanone → butanoic acid → butyl butanoate." (a) Identify the error in this pathway and explain why it is not achievable in Module 7. (b) Propose a valid alternative pathway starting from butan-1-ol that achieves butyl butanoate. Write equations and conditions for each step.

Q1 — Answer: C

Ethyl propanoate = ethyl group (C₂H₅-, from ethanol) + propanoate (from propanoic acid). Option C correctly identifies: ethanol from ethene (hydration: H₂O, H₃PO₄, 300°C) + propanoic acid from propan-1-ol (excess K₂Cr₂O₇/H₂SO₄, reflux); then esterification (conc. H₂SO₄, reflux, ⇌). Option B is wrong — esterification requires a CARBOXYLIC ACID, not an aldehyde. Option D names the wrong ester (ethanoic acid + propan-1-ol → propyl ethanoate, not ethyl propanoate).

Q2 — Answer: B

To stop oxidation at the aldehyde, the aldehyde must be removed from contact with oxidant before further oxidation occurs. Distillation achieves this — aldehyde has a lower boiling point than the parent alcohol, so it can be collected as it forms. Option A (reflux + excess) keeps aldehyde in contact with oxidant → over-oxidation to carboxylic acid. Option C (KMnO₄, reflux) is a stronger oxidant under reflux → carboxylic acid product. Option D (H₂SO₄, heat) gives dehydration → alkene, not oxidation.

Q3 — Answer: C

Reduction of a carboxylic acid to an alcohol is NOT in the Module 7 reaction set. Module 7 covers oxidation only — going up the oxidation ladder (alcohol → aldehyde → carboxylic acid). There is no Module 7 reagent for the reverse direction (reducing -COOH to -CH₂OH). Options A, B, D are all valid two-step pathways using known Module 7 conditions.

Q4 — Answer: A

1-chloropropane → propanoic acid in three steps: Step 1: NaOH(aq)/reflux → propan-1-ol (substitution). Step 2: K₂Cr₂O₇/H₂SO₄/distillation → propanal (mild oxidation to aldehyde). Step 3: K₂Cr₂O₇ excess/H₂SO₄/reflux → propanoic acid (further oxidation). Option C includes conc. H₂SO₄/reflux in step 3 which is esterification conditions, not oxidation. Option B starts with an oxidation step on a haloalkane which is not a Module 7 reaction.

Q5 — Answer: B

H₂SO₄ is a catalyst — it is consumed at the start and regenerated at the end. Mechanism: H⁺ from H₂SO₄ protonates the carbonyl oxygen of the carboxylic acid, activating it for nucleophilic attack by the alcohol's lone pair. At the end of the reaction, H⁺ is regenerated (no net consumption). Additionally, concentrated H₂SO₄ acts as a dehydrating agent — absorbing water produced in the reaction, reducing the concentration of water, and shifting the equilibrium right (Le Chatelier) to increase ester yield.

Q6 — Sample Answer (4 marks)

Step 1: CH₂=CH₂ + H₂O → CH₃CH₂OH. Conditions: H₂O (steam), H₃PO₄ catalyst, ~300°C, high pressure (~65 atm). Hydration of ethene by Markovnikov addition. Product: ethanol. [2 marks]
Step 2: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O (oxidise part of ethanol to ethanoic acid); then CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O. Conditions for oxidation: K₂Cr₂O₇/H₂SO₄ (excess), reflux. Conditions for esterification: conc. H₂SO₄ (catalyst), reflux, reversible (⇌). Product: ethyl ethanoate. [2 marks]

Q7 — Sample Answer (5 marks)

Step 1: CH₃CH₂CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaBr. Conditions: NaOH(aq), reflux. Product: butan-1-ol. [1 mark] Step 2: CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O. Conditions: K₂Cr₂O₇/H₂SO₄, distillation. Product: butanal. Distillation removes butanal (lower BP than butan-1-ol) as it forms, preventing excess oxidant from further oxidising it to butanoic acid. [1.5 marks] Step 3: CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH. Conditions: K₂Cr₂O₇/H₂SO₄ (excess), reflux. Product: butanoic acid. Reflux keeps butanal in contact with excess oxidant until complete conversion to carboxylic acid — no distillation needed here as full oxidation is desired. [1.5 marks] Step 4: CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂O. Conditions: conc. H₂SO₄ (catalyst), reflux, reversible (⇌). Product: butyl butanoate. [1 mark]

Q8 — Sample Answer (6 marks)

(a) Error: The step "butanone → butanoic acid" is not achievable in Module 7. Butanone (butan-2-one) is a ketone — the product of oxidising a secondary alcohol (butan-2-ol). Ketones cannot be further oxidised to carboxylic acids under Module 7 conditions because the carbonyl carbon has no C-H bond available for oxidation. The pathway reaches a dead end at butanone. [2 marks]
(b) Valid alternative from butan-1-ol: Step 1: CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O (K₂Cr₂O₇/H₂SO₄, distillation → butanal). Step 2: CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH (K₂Cr₂O₇ excess/H₂SO₄, reflux → butanoic acid). Step 3: CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂O (conc. H₂SO₄ cat., reflux, ⇌ → butyl butanoate). [4 marks — 1 per step equation + conditions + name of intermediate]

Revisit Your Think First Response

Ethene → ethyl ethanoate requires three steps: (1) hydration → ethanol; (2) full oxidation → ethanoic acid; (3) esterification with more ethanol → ethyl ethanoate. Did you identify all three steps? Did you notice that ethanol is used in two separate roles (as the starting material for oxidation AND as the alcohol for esterification)?

I can apply the four-step algorithm to construct multi-step organic synthesis pathways with full conditions

Previous: Organic Acids & Bases — pKa, Strength & Reactions ← Next: Organic Reactions Mastery — Conditions, Pathways & Band 6 Responses →

Module overview · Chemistry subject page · Next checkpoint · Module quiz